anonymous
  • anonymous
A car of mass 1.5 x 103 kg is hoisted on the large cylinder of a hydraulic press. The area of the large piston is 0.20 m2, and the area of the small piston is 0.015 m2. a.Calculate the magnitude of the force of the small piston needed to raise the car with slow speed on the large piston. b.Calculate the pressure, in pascals and kilopascals, in this hydraulic press.
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
@Michele_Laino
anonymous
  • anonymous
Hey there how are you?
anonymous
  • anonymous
I hope all is well with you;)

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Michele_Laino
  • Michele_Laino
yes! I'm fine! thanks! :)
Michele_Laino
  • Michele_Laino
here you have to keep in mind the principle of Pascal, namely the pressure of the fluid is the same on both pistons
anonymous
  • anonymous
Where water is used as multiplier of force in call cases.
anonymous
  • anonymous
Since for question a I know the area of small piston
anonymous
  • anonymous
I guess all I need is divide the mass to be lifted by the area of the piston?
Michele_Laino
  • Michele_Laino
more precisely are used special oils, instead of water. the pressure exerted by the fluid on larger piston is: \[\Large p = \frac{{Mg}}{S} = \frac{{1500 \cdot 9.81}}{{0.2}} = ...Pascals\]
anonymous
  • anonymous
Right... I needed to calculate the normal force acting on the object divided by the area of the piston.
Michele_Laino
  • Michele_Laino
that's right!
anonymous
  • anonymous
According to my calculator the pascals required on the piston is 73500 pascals
anonymous
  • anonymous
So that's considering if we only use the large piston right?
Michele_Laino
  • Michele_Laino
I got p = 73.6 kPa approximately
anonymous
  • anonymous
On the other hand if I use the smaller piston the answer would be slightly greater I think
anonymous
  • anonymous
I got p=981000 on my side for the smaller piston.
Michele_Laino
  • Michele_Laino
no, the pressure is the same through all the fluid. The force which is acting on the smaller piston, is perpendicular with respect to that piston, and its magnitude is: \[\Large F = p \times s = 73575 \cdot 0.015 = ...Newtons\]
anonymous
  • anonymous
Oh ok.. but why are we only using the pressure applied to the large piston? Is it because of the structure of hydraulic press?
Michele_Laino
  • Michele_Laino
no, we are using the pression of the fluid which is transmitted by the fluid itself, namely your hydraulic press can be represented as below: |dw:1439828282269:dw|
Michele_Laino
  • Michele_Laino
what changes are the magnitude of the forces applied to both pistons
Michele_Laino
  • Michele_Laino
magnitudes*
anonymous
  • anonymous
So basically a force is applied to one end and that results in other end
Michele_Laino
  • Michele_Laino
yes! we apply a little force and the hydraulic press multiply this little force
Michele_Laino
  • Michele_Laino
multiplies*
anonymous
  • anonymous
73575 Pa on the large piston.
anonymous
  • anonymous
I understand until that point
anonymous
  • anonymous
But then, isn't it like by 0.2/0.015 that the pressure is multiplied?
Michele_Laino
  • Michele_Laino
that pressure is acting inside all the fluid, so the pressure of 73575 is acting on both pistons
Michele_Laino
  • Michele_Laino
the pressure is the same!
anonymous
  • anonymous
Are you saying that fluid supplied from the inside the hydraulic press?
Michele_Laino
  • Michele_Laino
in your exercise, we apply a force whose magnitude is 1,104 Newtons, and we can rise a car whose weight is: 14,715 Newtons
Michele_Laino
  • Michele_Laino
yes! the fluid is working inside the hydraulic press
anonymous
  • anonymous
The force the small piston needs to lift the car right? so I need to apply force on the small piston to lift the car to the larger piston and force multiplies
Michele_Laino
  • Michele_Laino
correct! we can see a hydraulic press as a multiplier of force
anonymous
  • anonymous
So it's correct to say 73575 Pa* 0.015= pressure require to lift the car? The only thing I don't understand is why we multiply by 0.015 for a pressure calculated using 0.2. Isn't it like we have to multiply by 0.015/0.2 ?
Michele_Laino
  • Michele_Laino
pressure is 73575 Pa, and 73575*0.015 is a force
anonymous
  • anonymous
likewise 73575 Pa*0.2 is the force I need to apply to larger piston to do the same trick?
Michele_Laino
  • Michele_Laino
better is: 73575*0.2 is the force which is acting on the larger piston
anonymous
  • anonymous
wait a moment it's taking long so
anonymous
  • anonymous
that's my thanks for being patient with me XD
Michele_Laino
  • Michele_Laino
thanks! :)
anonymous
  • anonymous
I am trying my best to understand
Michele_Laino
  • Michele_Laino
ok! :)
anonymous
  • anonymous
You previously calculated the pressure applied to the large piston using the formula Mg/s=pascals. What if you used 0.015 instead of the 0.2? The overall pressure in the hydraulic press would then increase no?
anonymous
  • anonymous
According to your reasoning that pressure is the same across all points in a hydraulic press, then the pressure calculated using 0.015 instead of 0.2 would be greater...
Michele_Laino
  • Michele_Laino
I have used 0.2 since 0.2 is the area of the piston on which the weight of the car is acting on
Michele_Laino
  • Michele_Laino
I can not use 0.015, since it will be an error
Michele_Laino
  • Michele_Laino
|dw:1439829785001:dw|
Michele_Laino
  • Michele_Laino
what I can know is the pressure transmitted by the fluid through all the hydraulic press, and that pressure, by definition, is: \[\Large p = \frac{{Mg}}{S} = \frac{{1500 \cdot 9.81}}{{0.2}}\]
anonymous
  • anonymous
Oh I see then in terms of the larger piston the pressure acting on it is calculated by Mg/0.2. Right the pressure acting on the larger piston is accounted for by that. On the other hand, when I need to apply the pressure exerted by the car on larger piston, multiplying the previous pascals with 0.15 will be in accord with the principles of multiplication achieved by water.
anonymous
  • anonymous
So multiplying that number with 0.015 will give the force required to lift the car although at a slower rate.
anonymous
  • anonymous
Am I on the right line of reasoning?
Michele_Laino
  • Michele_Laino
yes! if I multiply the pressure by the area of the smaller piston, I get the force (magnitude) which I have to apply to that smaller piston, in order to raise the car
Michele_Laino
  • Michele_Laino
the concept is that pressure is the same in all points of the hydraulic press Whereas what is different are the magnitude of the forces which are acting on both pistons.
Michele_Laino
  • Michele_Laino
magnitudes*
anonymous
  • anonymous
Right.. so in that case water is used as multiplier of force to lift a really heavy piece of object.
Michele_Laino
  • Michele_Laino
that's right!
anonymous
  • anonymous
Yet the pressure acting in the water is same across all objects. Is it correct to say that water is dispersing the required force all through the water hence reducing the force required to lift the vehicle?
anonymous
  • anonymous
this is such a fascinating concept.. Quite the counterintuitive nature of our nature.
Michele_Laino
  • Michele_Laino
I think that it is correct to say that the effect of a force, is the production of the correspondig pressure inside the fluid which is contacting that force, and in turn the effect of a pressure of a fluid, is the force produced on all the surfaces which are contacting that fluid
Michele_Laino
  • Michele_Laino
and cause-effect, namely force-pressure, are descrined by the subsequent formula: \[\Large P = \frac{F}{S}\]
anonymous
  • anonymous
And depending on the area of surface that force induced by the object contacting the surface leading to space filled with liquids without leaks connected to another surface counteracting the force will be all dependent on the area upon which the respective forces are taking effect.
anonymous
  • anonymous
So within the given pressure, greater surface implies more force required to meet the same pressure, and less surface means less force required to meet that pressure.
Michele_Laino
  • Michele_Laino
no, greater is the surface and greater is the forces, and less is the surface and less is the force, since the ratio force/surface has to be constant. The magnitude of the resultant forces, or of the applied forces depends on the area of the surfaces contacting those forces
anonymous
  • anonymous
Yeah I realized I said the wrong
anonymous
  • anonymous
So I deleted it
Michele_Laino
  • Michele_Laino
ok! your last reply is correct!
anonymous
  • anonymous
Isn't this counterintuitive?
anonymous
  • anonymous
Well, I guess the expanse of liquid present underneath the surface with greater area will make sense that corresponding force required induce such pressure is indeed greater
anonymous
  • anonymous
On the other hand smaller surface with smaller liquid underneath would make it less force required the meet the pressure although distributed throughout all the fluid bodies.
Michele_Laino
  • Michele_Laino
I don't think, I think that it is intuitive, the problem is that we have no experience, in our daily life, about expansion of fluid
Michele_Laino
  • Michele_Laino
yes! exact!
anonymous
  • anonymous
Yes! Yes! lack of experience with interaction of liquid makes it harder for us to visualize.
Michele_Laino
  • Michele_Laino
yes! and the best way is to use our imagination. As Einstein said: "Imagination is more important than knowledge"
anonymous
  • anonymous
On the other hand, studying physics makes you capable of physically achieving anything if you know how to properly coordinate your muscles to execute the proper force required to achieve that phenomena.
anonymous
  • anonymous
I think, in my personal opinion, all children should be taught the principles of physics so as to excel as well as increase "efficiency" in life. So much are wasted by ignorance
anonymous
  • anonymous
I think too, that physics offers us a exemplary thinking mechanism.
anonymous
  • anonymous
Also physics extends our lifespan
anonymous
  • anonymous
By creating new neurons in your brain
Michele_Laino
  • Michele_Laino
I think that there is a time for study physics, what is important for children is play, since by means of play, all children can collect those experience which will allow to think in terms of our imagination
anonymous
  • anonymous
True, from those natural experience will help us to relate physics to our past times. Richer the past times, greater the utility for physics
Michele_Laino
  • Michele_Laino
correct!
anonymous
  • anonymous
And physics makes us nostalgic of our past memories.
Michele_Laino
  • Michele_Laino
yes!
anonymous
  • anonymous
And ultimately makes life worthwhile. Physics makes all my struggles disappear.
Michele_Laino
  • Michele_Laino
throw a rock in the air, or running through a meadow, are all important things
anonymous
  • anonymous
Physics is the best prescription for eliminating societal prejudice and personal stigmas for it provides us more objective insight upon which to rely all of the issues we encounter in life. This makes it easy for us to say that someone is wrong in saying something or the applications of medical physics such as how our neutrons interact would invent better ways to learn
anonymous
  • anonymous
it's all physics after all our world is made of. And contemporary education has so far failed to quantify the most effective means by which to learn, hence resulting in "learning something until it finally sinks in" or just relying on our intuition rather than physics of neurology.
anonymous
  • anonymous
that's why I love physics .
Michele_Laino
  • Michele_Laino
:)
anonymous
  • anonymous
I bet you were on the same road when you were younger
Michele_Laino
  • Michele_Laino
not exactly, when I was younger, I had the concern, to give a logical accommodation to everything around me
anonymous
  • anonymous
Oh I see, trying to explain mathematically everything around you o
Michele_Laino
  • Michele_Laino
yes! a logical or rational explanation
Michele_Laino
  • Michele_Laino
ok! when you need help in physics, keep in mind that I always will help you! :)
anonymous
  • anonymous
Thanks so much Michele!
Michele_Laino
  • Michele_Laino
:)
anonymous
  • anonymous
Do you mind proof reading another application from this question that I learned?
anonymous
  • anonymous
2. I calculated the pressure acting on the floor when standing on both feet. The areas of my shoes were approximated to be 15cm^2 each, making ti 30cm^2. Given that I weigh around 58kg, and pressure is calculated as P=Mg/s (surface area) P=58kg*9.8m/s^2/0.003m^3=189466.667 The pressure acting on the floor when standing on both feet is 1.9*10^5 Pa
anonymous
  • anonymous
According to your P=Mg/s I think my reasoning above seems pretty good
Michele_Laino
  • Michele_Laino
yes! Your reasoning is correct!

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